P4 Stress and Strain Dr. A.B. Zavatsky HT08 Lecture 5 Plane Stress Transformation Equations Stress elements and plane stress. Stresses on inclined sections. Transformation equations. Principal stresses, angles, and planes. Maximum shear stress. 1 Normal and shear stresses on inclined sections To obtain a complete picture of the stresses in a bar, we must consider the stresses acting on an “inclined” (as opposed to a “normal”) section through the bar. Inclined section Normal section P P Because the stresses are the same throughout the entire bar, the stresses on the sections are uniformly distributed. P Inclined section P Normal section 2 y P N θ x P V The force P can be resolved into components: Normal force N perpendicular to the inclined plane, N = P cos θ Shear force V tangential to the inclined plane V = P sin θ If we know the areas on which the forces act, we can calculate the associated stresses. y area A y σθ x area (A / cos θ) τθ area A x area (A / cos θ) 3 y τθ σθ θ x σθ = Force N P cos θ P = = = cos 2θ Area Area A / cos θ A σ θ = σ x cos 2θ = σx (1+ cos 2θ ) 2 σmax = σx occurs when θ = 0° τθ = −V − P sin θ Force P = = = − sin θ cos θ Area Area A / cos θ A τ θ = −σ x sin θ cos θ = − σx (sin 2θ ) 2 τmax = ± σx/2 occurs when θ = -/+ 45° 4 Introduction to stress elements Stress elements are a useful way to represent stresses acting at some point on a body. Isolate a small element and show stresses acting on all faces. Dimensions are “infinitesimal”, but are drawn to a large scale. y P σx = P / A x Area A y σx y σx = P/A x σx σx = P/A x z 5 Maximum stresses on a bar in tension P P a σx σx = σmax = P / A a Maximum normal stress, Zero shear stress 6 Maximum stresses on a bar in tension P P a b σx/2 θ = 45° τmax = σx/2 σx/2 b Maximum shear stress, Non-zero normal stress 7 Stress Elements and Plane Stress When working with stress elements, keep in mind that only one intrinsic state of stress exists at a point in a stressed body, regardless of the orientation of the element used to portray the state of stress. We are really just rotating axes to represent stresses in a new coordinate system. y1 y x1 σx θ σx x 8 y σy τyz τxy τzy σx Normal stresses σx, σy, σz (tension is positive) τyx τzx τxz σz σx x Shear stresses τxy = τyx, τxz = τzx, τyz = τzy z σy Sign convention for τab Subscript a indicates the “face” on which the stress acts (positive x “face” is perpendicular to the positive x direction) Subscript b indicates the direction in which the stress acts Strictly σx = σxx, σy = σyy, σz = σzz 9 When an element is in plane stress in the xy plane, only the x and y faces are subjected to stresses (σz = 0 and τzx = τxz = τzy = τyz = 0). Such an element could be located on the free surface of a body (no stresses acting on the free surface). Plane stress element in 2D σx, σy τxy = τyx σz = 0 σy τyx τxy y x σx σx τxy τyx σy 10 Stresses on Inclined Sections y1 σy τxy x σx σy1 τy1x1 τyx y τyx τx1y1 x1 σx1 θ σx σx1 τxy y τx1y1 x τy1x1 σy1 The stress system is known in terms of coordinate system xy. We want to find the stresses in terms of the rotated coordinate system x1y1. Why? A material may yield or fail at the maximum value of σ or τ. This value may occur at some angle other than θ = 0. (Remember that for uniaxial tension the maximum shear stress occurred when θ = 45 degrees. ) 11 Transformation Equations Forces Stresses y y y1 y1 θ σx x1 τx1y1 σx1 θ θ σxA σx1 A sec θ x τxy τx1y1 A sec θ x1 θ x τxy A τyx τyx A tan θ σy Forces can be found from stresses if the area on which the stresses act is known. Force components can then be summed. σy A tan θ Left face has area A. Bottom face has area A tan θ. Inclined face has area A sec θ. 12 y y1 θ σxA τx1y1 A sec θ σx1 A sec θ x1 θ x τxy A τyx A tan θ σy A tan θ Sum forces in the x1 direction : σ x1 A sec θ − (σ x A) cos θ − (τ xy A)sin θ − (σ y A tan θ )sin θ − (τ yx A tan θ )cos θ = 0 Sum forces in the y1 direction : τ x1 y1 A sec θ + (σ x A)sin θ − (τ xy A)cos θ − (σ y A tan θ )cos θ − (τ yx A tan θ )sin θ = 0 Using τ xy = τ yx and simplifying gives : σ x1 = σ x cos 2 θ + σ y sin 2 θ + 2 τ xy sin θ cos θ τ x1 y1 = − (σ x − σ y ) sin θ cos θ + τ xy (cos 2 θ − sin 2 θ ) 13 Using the following trigonometric identities cos 2 θ = 1 + cos 2θ 2 sin 2 θ = 1 − cos 2θ 2 sin θ cos θ = sin 2θ 2 gives the transformation equations for plane stress : σ x +σ y σ x −σ y σ x1 = + cos 2θ + τ xy sin 2θ 2 2 HLT, page 108 ( σ x −σ y ) τ x1 y1 = − sin 2θ + τ xy cos 2θ 2 For stresses on the y1 face, substitute θ + 90° for θ : σ y1 = σx +σ y 2 − σ x −σ y 2 cos 2θ − τ xy sin 2θ Summing the expressions for x1 and y1 gives : σ x1 + σ y1 = σ x + σ y Can be used to find σy1, instead of eqn above. 14 Example: The state of plane stress at a point is represented by the stress element below. Determine the stresses acting on an element oriented 30° clockwise with respect to the original element. 50 MPa y 80 MPa 80 MPa x 25 MPa 50 MPa Define the stresses in terms of the established sign convention: σx = -80 MPa σy = 50 MPa τxy = -25 MPa We need to find σx1, σy1, and τx1y1 when θ = -30°. Substitute numerical values into the transformation equations: σ x1 = σx +σ y + σ x −σ y cos 2θ + τ xy sin 2θ 2 2 − 80 + 50 − 80 − 50 σ x1 = + cos 2(− 30°) + (− 25)sin 2(− 30°) = −25.9 MPa 2 2 15 σ y1 = σ x +σ y − σ x −σ y cos 2θ − τ xy sin 2θ 2 2 − 80 + 50 − 80 − 50 σ y1 = − cos 2(− 30°) − (− 25)sin 2(− 30°) = −4.15 MPa 2 2 σ x −σ y sin 2θ + τ xy cos 2θ τ x1 y1 = − 2 (− 80 − 50) sin 2(− 30°) + (− 25) cos 2(− 30°) = −68.8 MPa τ x1 y1 = − 2 ( ) y 25.8 MPa y1 4.15 MPa +60° o -30 x Note that σy1 could also be obtained (a) by substituting +60° into the equation for σx1 or (b) by using the equation σx + σy = σx1 + σy1 25.8 MPa x1 4.15 MPa 68.8 MPa (from Hibbeler, Ex. 15.2) 16 Principal Stresses The maximum and minimum normal stresses (σ1 and σ2) are known as the principal stresses. To find the principal stresses, we must differentiate the transformation equations. σ x1 = σx +σ y 2 σ x −σ y + σ x −σ y 2 cos 2θ + τ xy sin 2θ dσ x1 (− 2 sin 2θ ) + τ xy (2 cos 2θ ) = 0 = dθ 2 dσ x1 = − (σ x − σ y ) sin 2θ + 2τ xy cos 2θ = 0 dθ 2τ xy θp are principal angles associated with tan 2θ p = σ x −σ y the principal stresses (HLT, page 108) There are two values of 2θp in the range 0-360°, with values differing by 180°. There are two values of θp in the range 0-180°, with values differing by 90°. So, the planes on which the principal stresses act are mutually perpendicular. 17 We can now solve for the principal stresses by substituting for θp in the stress transformation equation for σx1. This tells us which principal stress is associated with which principal angle. 2τ xy tan 2θ p = σ x −σ y σ x1 = σ x +σ y 2 + σ x −σ y 2 cos 2θ + τ xy sin 2θ ⎛σ x −σ y ⎞ R 2 = ⎜⎜ ⎝ R τxy 2θp sin 2θ p = (σx – σy) / 2 σ1 = σx +σ y 2 + cos 2θ p = 2 2 ⎟⎟ + τ xy 2 ⎠ σ x −σ y τ xy 2R R σ x −σ y ⎛σ x −σ y ⎞ 2 ⎜⎜ ⎝ 2R ⎛ τ xy ⎞ ⎟⎟ + τ xy ⎜⎜ ⎟⎟ ⎠ ⎝ R ⎠ 18 Substituting for R and re-arranging gives the larger of the two principal stresses: σ1 = σ x +σ y 2 ⎛σ x −σ y ⎜⎜ ⎝ 2 + 2 ⎞ ⎟⎟ + τ xy 2 ⎠ To find the smaller principal stress, use σ1 + σ2 = σx + σy. σ 2 = σ x + σ y − σ1 = σ x +σ y 2 ⎛σ x −σ y ⎜⎜ ⎝ 2 − 2 ⎞ ⎟⎟ + τ xy 2 ⎠ These equations can be combined to give: σ 1, 2 = σ x +σ y 2 ± ⎛ σ x −σ y ⎞ ⎜⎜ ⎝ 2 2 ⎟⎟ + τ xy 2 ⎠ Principal stresses (HLT page 108) To find out which principal stress goes with which principal angle, we could use the equations for sin θp and cos θp or for σx1. 19 The planes on which the principal stresses act are called the principal planes. What shear stresses act on the principal planes? Compare the equations for τ x1 y1 = 0 and dσ x1 dθ = 0 τ x1 y1 = ( σ − x −σ y ) sin 2θ + τ xy cos 2θ = 0 2 − (σ x − σ y ) sin 2θ + 2τ xy cos 2θ = 0 dσ x1 = − (σ x − σ y ) sin 2θ + 2τ xy cos 2θ = 0 dθ Solving either equation gives the same expression for tan 2θp Hence, the shear stresses are zero on the principal planes. 20 y σy τyx σ2 τxy y x σx θp2 σx σ1 θp1 x σ1 τxy τyx σ2 σy Principal Stresses σ 1, 2 = σ x +σ y 2 ± ⎛ σ x −σ y ⎞ ⎜⎜ ⎝ 2 2 ⎟⎟ + τ xy 2 ⎠ Principal Angles defining the Principal Planes tan 2θ p = 2τ xy σ x −σ y 21 Example: The state of plane stress at a point is represented by the stress element below. Determine the principal stresses and draw the corresponding stress element. 50 MPa y 80 MPa 80 MPa x Define the stresses in terms of the established sign convention: σx = -80 MPa σy = 50 MPa τxy = -25 MPa 25 MPa 50 MPa σ 1, 2 = σ 1, 2 σx +σ y 2 ± − 80 + 50 = ± 2 σ 1 = 54.6 MPa ⎛σ x −σ y ⎜⎜ ⎝ 2 2 ⎞ ⎟⎟ + τ xy 2 ⎠ ⎛ − 80 − 50 ⎞ 2 ⎜ ⎟ + (− 25) = −15 ± 69.6 2 ⎝ ⎠ σ 2 = −84.6 MPa 2 22 tan 2θ p = tan 2θ p = 2τ xy y σx −σ y 54.6 MPa 2(− 25) = 0.3846 − 80 − 50 84.6 MPa o 100.5 84.6 MPa o 10.5 x 2θ p = 21.0° and 21.0 + 180° θ p = 10.5°, 100.5° 54.6 MPa But we must check which angle goes with which principal stress. σ x1 = σx +σ y + σ x −σ y cos 2θ + τ xy sin 2θ 2 2 − 80 + 50 − 80 − 50 σ x1 = + cos 2(10.5°) + (− 25)sin 2(10.5°) = −84.6 MPa 2 2 σ1 = 54.6 MPa with θp1 = 100.5° σ2 = -84.6 MPa with θp2 = 10.5° 23 The two principal stresses determined so far are the principal stresses in the xy plane. But … remember that the stress element is 3D, so there are always three principal stresses. yp y σ2 σy τ σ1 τ σx σx τ x σ1 σ3 = 0 τ σy σx, σy, τxy = τyx = τ xp zp σ2 σ1, σ2, σ3 = 0 Usually we take σ1 > σ2 > σ3. Since principal stresses can be compressive as well as tensile, σ3 could be a negative (compressive) stress, rather than the zero stress. 24 Maximum Shear Stress To find the maximum shear stress, we must differentiate the transformation equation for shear. ( σ x −σ y ) τ x1 y1 = − sin 2θ + τ xy cos 2θ 2 dτ x1 y1 = − (σ x − σ y )cos 2θ − 2τ xy sin 2θ = 0 dθ ⎛σ x −σ y ⎞ ⎟ tan 2θ s = − ⎜ ⎜ 2τ ⎟ xy ⎝ ⎠ There are two values of 2θs in the range 0-360°, with values differing by 180°. There are two values of θs in the range 0-180°, with values differing by 90°. So, the planes on which the maximum shear stresses act are mutually perpendicular. Because shear stresses on perpendicular planes have equal magnitudes, the maximum positive and negative shear stresses differ only in sign. 25 We can now solve for the maximum shear stress by substituting for θs in the stress transformation equation for τx1y1. (σx – σy) / 2 ⎛σ x −σ y ⎞ ⎟ tan 2θ s = − ⎜ ⎟ ⎜ 2τ xy ⎝ ⎠ ( σ x −σ y ) τ x1 y1 = − sin 2θ + τ xy cos 2θ 2 R 2θs τxy τ max = ⎛σ x −σ y ⎜⎜ 2 ⎝ ⎛σ x −σ y ⎞ R 2 = ⎜⎜ ⎝ cos 2θ s = 2 ⎟⎟ + τ xy 2 ⎠ 2 τ xy R sin 2θ s = − σ x −σ y 2R 2 ⎞ ⎟⎟ + τ xy 2 ⎠ τ min = − τ max 26 Use equations for sin θs and cos θs or τx1y1 to find out which face has the positive shear stress and which the negative. What normal stresses act on the planes with maximum shear stress? Substitute for θs in the equations for σx1 and σy1 to get σ x1 = σ y1 = 2 =σs y σy σs τmax τyx τxy y x σx σx +σ y τyx σy σs θs σx σs τxy τmax τmax x τmax σs 27 Example: The state of plane stress at a point is represented by the stress element below. Determine the maximum shear stresses and draw the corresponding stress element. 50 MPa y 80 MPa 80 MPa x Define the stresses in terms of the established sign convention: σx = -80 MPa σy = 50 MPa τxy = -25 MPa 25 MPa 50 MPa τ max = ⎛σ x −σ y ⎞ ⎜⎜ ⎝ 2 2 ⎟⎟ + τ xy ⎠ ⎛ − 80 − 50 ⎞ 2 ⎜ ⎟ + (− 25) = 69.6 MPa 2 ⎝ ⎠ 2 τ max = 2 σs = σx +σ y 2 − 80 + 50 σs = = −15 MPa 2 28 ⎛ σ x −σ y ⎞ ⎟ = − ⎛⎜ − 80 − 50 ⎞⎟ = −2.6 tan 2θ s = − ⎜ ⎜ 2 (− 25) ⎟ ⎜ 2τ ⎟ ⎠ ⎝ xy ⎝ ⎠ 2θ s = −69.0° and − 69.0 + 180° y 15 MPa 15 MPa θ s = −34.5°, 55.5° 55.5 But we must check which angle goes with which shear stress. τ x1 y1 = ( σ − x −σ y ) o x -34.5o 15 MPa sin 2θ + τ xy cos 2θ 15 MPa 69.6 MPa 2 (− 80 − 50) sin 2(− 34.5) + (− 25) cos 2(− 34.5) = −69.6 MPa τ x1 y1 = − 2 τmax = 69.6 MPa with θsmax = 55.5° τmin = -69.6 MPa with θsmin = -34.5° 29 Finally, we can ask how the principal stresses and maximum shear stresses are related and how the principal angles and maximum shear angles are related. σ 1, 2 = σx +σ y 2 ± ⎛σ x −σ y ⎜⎜ ⎝ 2 2 ⎞ ⎟⎟ + τ xy 2 ⎠ ⎛σ x −σ y ⎞ ⎜⎜ ⎟⎟ + τ xy 2 ⎝ 2 ⎠ 2 σ1 − σ 2 = 2 σ 1 − σ 2 = 2τ max σ −σ2 τ max = 1 2 ⎛σ x −σ y ⎞ 2τ xy ⎟ tan 2θ s = − ⎜ tan 2θ p = ⎜ 2τ ⎟ σ x −σ y xy ⎝ ⎠ −1 tan 2θ s = = − cot 2θ p tan 2θ p 30 tan 2θ s + cot 2θ p = 0 cos 2θ p sin 2θ s + =0 cos 2θ s sin 2θ p sin 2θ s sin 2θ p + cos 2θ s cos 2θ p = 0 cos (2θ s − 2θ p ) = 0 2θ s − 2θ p = ± 90° θ s − θ p = ± 45° θ s = θ p ± 45° So, the planes of maximum shear stress (θs) occur at 45° to the principal planes (θp). 31 Original Problem Principal Stresses y 50 MPa 54.6 MPa y 80 MPa 84.6 MPa o 100.5 80 MPa x 84.6 MPa o 10.5 x 25 MPa 50 MPa 54.6 MPa σx = -80, σy = 50, τxy = 25 σ1 = 54.6, σ2 = 0, σ3 = -84.6 Maximum Shear y 15 MPa 15 MPa τ max = 55.5 o x -34.5o 15 MPa 15 MPa τ max τ max σ1 − σ 2 θ s = θ p ± 45° 2 θ = 10.5° ± 45° 54.6 − (− 84.6 ) s = θ s = −34.5°, 55.5° 2 = 69.6 MPa 69.6 MPa τmax = 69.6, σs = -15 32
